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I do think I'm wording this somewhat incorrect so I'll do my best to explain what my goal is. I'm trying to create a summation that adds $n$ values repeatedly until $m$. So it'd look something like this: $(0) + (0+1)+ (0+1+2) + ... + (0+1+2+...+m)$

I've tried setting up the summation though I'm not really sure how it would look when it's set up something like

$$ \[ \sum_{n=0}^{m} (n+...) \] $$

This may be basic but I'm unsure what'd the inside of the summation would look like.

Nom
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  • Well, you can nest: $\sum_{\omega=0}^m (\sum_{k=0}^\omega k)$ can do what you ask. On the other hand we can algebraically solve that $\sum_{\omega=0}^m (\sum_{k=0}^\omega k)=\sum_{\omega=0}^m \frac {\omega(\omega + 1}2=$$\sum_{\omega=0}^m[\frac {\omega^2}2 + \frac \omega2] =\frac 12(\sum_{\omega}^m \omega^2 + \sum_{\omega=0}^m \omega) = \frac 12(\frac {m(m+1)(2m+1)}6 + \frac {m(m+1)}2)=\frac {m(m+1)(2m+4) }{12}$. – fleablood Oct 18 '22 at 04:03

2 Answers2

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The $j$th term in the expression you want is $$ \sum_{i=0}^j i; $$ and you want to write down all those terms as $j$ runs from $0$ to $m$, which can be done with another summation: $$ \sum_{j=0}^m \sum_{i=0}^j i. $$ (By the way, a formula for the sum turns out to be $\frac16m(m+1)(2m+1)$. But it sounds like that's a secondary concern for you.)

Greg Martin
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$(0) + (0+1)+ (0+1+2) + ... + (0+1+2+...+m)$

Alternative expression, by analysis:

  • $(0)$ is being added $(m+1)$ times.

  • $(1)$ is being added $(m)$ times.

  • $(2)$ is being added $(m-1)$ times.

  • $\cdots$.

  • $(m)$ is being added $(1)$ time.

So, you want

$$\sum_{i=0}^m [f(i) \times i] ~: ~f(i) = (m+1-i).$$

This can be expressed as

$$\sum_{i=0}^m [(m+1-i) \times i].$$

user2661923
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